| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Solving linear systems using matrices |
| Difficulty | Standard +0.3 This is a straightforward applied problem requiring students to set up three linear equations from a word problem and solve using matrices. The setup is mechanical (translating constraints into equations), and the matrix solution is routine Gaussian elimination or inverse matrix method. Slightly easier than average as it's a standard 'real-world' application with no conceptual subtlety beyond basic equation formation. |
| Spec | 4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=\) value of savings account, \(y=\) value of property bond account, \(z=\) value of share dealing account | M1 | Attempts to set up 3 equations with 3 unknowns |
| \(x+y+z=5000\); \(x+400=y\); \(0.015x+0.035y-0.025z=79\) | A1 | At least 2 equations correct with appropriate variables defined |
| \(\begin{pmatrix}1&1&1\\1&-1&0\\0.015&0.035&-0.025\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5000\\-400\\79\end{pmatrix}\) | M1 | Sets up a matrix equation of the form \(A\mathbf{x}=\mathbf{b}\) |
| Correct matrix equation | A1 | |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1&1&1\\1&-1&0\\0.015&0.035&-0.025\end{pmatrix}^{-1}\begin{pmatrix}5000\\-400\\79\end{pmatrix}\) | M1 | Depends on previous M mark. Applies \(\mathbf{A}^{-1}\) and obtains at least one value of \(x\), \(y\) or \(z\) |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1800\\2200\\1000\end{pmatrix}\) | A1 | Correct answer |
| Tyler invested £1800 in savings account, £2200 in property bond account and £1000 in share dealing account | A1ft | Correct follow through answer in context |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=$ value of savings account, $y=$ value of property bond account, $z=$ value of share dealing account | M1 | Attempts to set up 3 equations with 3 unknowns |
| $x+y+z=5000$; $x+400=y$; $0.015x+0.035y-0.025z=79$ | A1 | At least 2 equations correct with appropriate variables defined |
| $\begin{pmatrix}1&1&1\\1&-1&0\\0.015&0.035&-0.025\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5000\\-400\\79\end{pmatrix}$ | M1 | Sets up a matrix equation of the form $A\mathbf{x}=\mathbf{b}$ |
| Correct matrix equation | A1 | |
| $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1&1&1\\1&-1&0\\0.015&0.035&-0.025\end{pmatrix}^{-1}\begin{pmatrix}5000\\-400\\79\end{pmatrix}$ | M1 | Depends on previous M mark. Applies $\mathbf{A}^{-1}$ and obtains at least one value of $x$, $y$ or $z$ |
| $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1800\\2200\\1000\end{pmatrix}$ | A1 | Correct answer |
| Tyler invested £1800 in savings account, £2200 in property bond account and £1000 in share dealing account | A1ft | Correct follow through answer in context |
---\begin{enumerate}
\item Tyler invested a total of $\pounds 5000$ across three different accounts; a savings account, a property bond account and a share dealing account.
\end{enumerate}
Tyler invested $\pounds 400$ more in the property bond account than in the savings account.\\
After one year
\begin{itemize}
\item the savings account had increased in value by $1.5 \%$
\item the property bond account had increased in value by $3.5 \%$
\item the share dealing account had decreased in value by $2.5 \%$
\item the total value across Tyler's three accounts had increased by $\pounds 79$
\end{itemize}
Form and solve a matrix equation to find out how much money was invested by Tyler in each account.
\hfill \mbox{\textit{Edexcel CP AS Q3 [7]}}